Chapter 7.3, Problem 30PPS

To determine

To graph: the triangle and prove that $\Delta XYZ\sim \Delta WYV$ .

Answer to Problem 30PPS

The two-column theorem is proved.

Explanation of Solution

Given:

  $\Delta XYZ\text{ and }\Delta WYV$ has vertices $X\left(-1,-9\right),Y\left(5,3\right)Z\left(-1,6\right),W\left(-1,5\right)V\left(1,5\right)$ .

Calculation:

Graph of the triangle is shown below:

  

Using distance formula to find the length of the sides.

As the distance formula is $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

For $\Delta XYZ$ ,

  $\begin{array}{l}XY=\sqrt{{\left(5-\left(-1\right)\right)}^2+{\left(3-\left(-9\right)\right)}^2}=\sqrt{{\left(6\right)}^2+{\left(12\right)}^2}=\sqrt{180}=6\sqrt{5}\\ YZ=\sqrt{{\left(-1-5\right)}^2+{\left(6-3\right)}^2}=\sqrt{{\left(-6\right)}^2+{\left(3\right)}^2}=\sqrt{45}=3\sqrt{5}\\ ZX=\sqrt{{\left(-1-\left(-1\right)\right)}^2+{\left(-9-6\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(15\right)}^2}=\sqrt{225}=15\end{array}$

For $\Delta WYV$ ,

  $\begin{array}{l}WY=\sqrt{{\left(5-1\right)}^2+{\left(3-\left(-5\right)\right)}^2}=\sqrt{{\left(4\right)}^2+{\left(8\right)}^2}=\sqrt{80}=4\sqrt{5}\\ YV=\sqrt{{\left(1-5\right)}^2+{\left(5-3\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(2\right)}^2}=\sqrt{20}=2\sqrt{5}\\ VW=\sqrt{{\left(1-1\right)}^2+{\left(-5-5\right)}^2}=\sqrt{{\left(0\right)}^2+{\left(-10\right)}^2}=\sqrt{100}=10\end{array}$

Now, check the proporsanality of the corresponding sides. Then,

  $\frac{XY}{WY}=\frac{6\sqrt{5}}{4\sqrt{5}}=\frac{3}{2};\frac{YZ}{YV}=\frac{3\sqrt{5}}{2\sqrt{5}}=\frac{3}{2};\frac{ZX}{VW}=\frac{15}{10}=\frac{3}{2}$

Thus, by SSS- similarity, the given triangle is similar. i.e. $\Delta XYZ\sim \Delta WYV$

Hence, the two-column theorem is proved.